Prove that the curves y2 = 4x and
x2 + y2–6x + 1 = 0 touch each other at the point (1, 2).
Given:
Curves y2 = 4x ...(1)
& x2 + y2 – 6x + 1 = 0 ...(2)
∴The point of intersection of two curves is (1,2)
First curve is y2 = 4x
Differentiating above w.r.t x,
⇒ 2y. = 4
⇒ y. = 2
⇒ m1 ...(3)
Second curve is x2 + y2 – 6x + 1 = 0
⇒ 2x + 2y.6 – 0 = 0
⇒ x + y.3 = 0
⇒ y. = 3 – x
...(4)
At (1,2),we have,
m1
⇒
m1 = 1
At (1,2),we have,
⇒ m2
⇒
⇒ m2 = 1
Clearly, m1 = m2 = 1 at (1,2)
So, given curve touch each other at (1,2)